1. Interpret the circuit

Since the battery has emf $E=2\,\text{V}$ and internal resistance $r=\dfrac{2}{3}\,\Omega$, the total power consumed in the entire circuit is the power delivered by the battery to all resistive elements including its internal resistance.

To compute that, we need the total current:

$I=\frac{E}{R_{\text{ext}}+r}$

From the given circuit, the external equivalent resistance is

$R_{\text{ext}}=\frac{2}{3}\,\Omega$

Hence total resistance of the whole circuit is

$R_{\text{total}}=R_{\text{ext}}+r=\frac{2}{3}+\frac{2}{3}=\frac{4}{3}\,\Omega$

2. Find the circuit current

$I=\frac{E}{R_{\text{total}}}=\frac{2}{4/3}=\frac{3}{2}\,\text{A}$

3. Find total power consumption

Total power consumed in the entire circuit is

$P=EI$

So,

$P=2\times \frac{3}{2}=3\,\text{W}$

(Equivalently, $P=I^2R_{\text{total}}=(\frac{3}{2})^2\cdot \frac{4}{3}=3\,\text{W}$.)

4. Final answer

$\boxed{3\,\text{W}}$

5. Comparison with stored answer

Stored correct answer = $3$

Our derived answer also equals $3$, so they agree.